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Proof step 17 from proof newrefs-two (previous step)

∃ sk. newref(st[newref(st) -- 0 ', noval]) -- sk ∈ st ⊦

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 8:
⊦ r -- sk ∈ st → r ∈ st
Proof step 16:
newref(st[newref(st) -- 0 ', noval]) -- sk ∈ st, newref(st[newref(st) -- 0 ', noval]) ∈ st ⊦

Rule argument

[emptyarg]

Additional information

[[CONCRETESIMPRULES : ()]]